# Techniques for Computing Polynomial Roots

2023-04-06 00:51:38 - Grace Browns Grace Browns has been a lifestyle, fashion, and beauty writer for over 5 years, and she currently serves as a senior editor at 422346.com.

### Results of Education

• Use the Remainder Theorem to assess a polynomial.
• To locate rational zeros, just apply the Rational Zero Theorem.
• A polynomial equation can be solved by applying the Factor Theorem.
• In order to determine a polynomial function's zeros, synthetic division can be used.
• In order to determine the complex zeros of a polynomial function, one can turn to the Fundamental Theorem of Algebra.
• Find polynomials that contain given zeros by applying the Linear Factorization Theorem
• To find out how many real zeros a polynomial function can have, apply Descartes' Rule of Signs.
• Find practical uses for polynomial equations and their solutions.

A new bakery has plain and themed sheet cakes for all your celebration needs. The cake shop requests that a miniature cake has a volume of 351 cubic inches. The cake takes the form of a rectangular cube. They requested that the cake's length be four inches longer than its width, and that its height be one-third that of its width. How big of a cake pan should I get?

The volume of the cake can be calculated by writing a cubic function and solving a cubic equation. Several methods for formulating polynomial functions and solving polynomial equations will be covered below.

## Polynomial Function Analysis Theorems

How to divide polynomials was covered in the previous section. The Remainder Theorem allows us to evaluate polynomials by dividing them. If the polynomial is divided by x – k, the remainder may be found quickly by evaluating the polynomial function at k, that is, f(k) Let's work through the theorem's proof.

Remember that for any polynomial f(x) and any non-zero polynomial d(x) where d(x) has degree less than or equal to f(x), there exists a unique polynomial q(x) and r(x) such that f(x)d(x).

F(x,y)=D(x,y),Q(x,y), R(x,y) is a common trig identity.

In the case where d(x) = x - k, the form

When writing in LaTeX, one can write: fleft(xright)=left(x-kright)qleft(xright) r.

The constant r will be the remainder when dividing by the linear divisor x - k. And if we check this for the case of x = k, we get

For example, $beginarraylfleft(kright)=left(k-k-right)qleft(k-right) rhfill textfleft(k-right)=0cdot qleft(k-right) rhfill textfleft(k-right)=rhfill text To rephrase, f(k) equals the remainder after dividing f(x) by (x - k). The remainder obtained by dividing a polynomial [latex]fleft(xright)$ by $x - kright$ is equal to the value of the polynomial $fleft(kright] in terms of x. ### The Remainder Theorem is used to determine the value of the function f(x) at the point x=k for a given polynomial f. 1. Subtract x-k from the polynomial using synthetic division. 2. The value obtained after subtracting (f-left) and (k-right) is the remainder. At [latex]x=2$, $fleft(xright)=6x4-x3-15x2 2x - 7$ can be calculated using the Remainder Theorem.

The Remainder Theorem can be used to determine the value of the expression $fleft(xright)=2x5 4x4-3x3 8x2 7$.
the value $x=3$

To determine if a rational number is a zero for a given polynomial, the Remainder Theorem can be used, but first we need a set of rational numbers to use it on. By comparing the factors of the constant term to those of the leading coefficient of the polynomial, the Rational Zero Theorem allows us to eliminate some of the possible rational zeros.

Think of the quadratic function y=x2+x2+x3+x3+x3+x3+x3+x3+x3+x3+x3+x3+x3+x3+x3+x3+x3+x3

There are associated factors with these 0s. The original quadratic function can be reconstructed by setting all of the factors to 0.

The original rational roots (2 and 3) are reflected in two of the factors of the constant term (6). Similarly, the original rational roots' denominators, 5 and 4, are two of the factors in the leading coefficient, 20.

It follows that both the numerators and denominators of all rational roots are factors of the constant term and the leading coefficient. In essence, the Rational Zero Theorem provides us with a set of potential rational zeros in this way.

If the coefficients of a polynomial $fleft(xright)=a_nxn a_n - 1xn a_1x a_0$ are integers, then the Rational Zero Theorem holds. If p is a factor of the constant term $a__0$ and q is a factor of the leading coefficient $a__n$, then every rational zero of $fleft(xright)$ takes the form $fracpq$.

When the leading coefficient is 1, the factors of the constant term are the rational zeros.

### Find rational zeros of the polynomial function f(x) by applying the Rational Zero Theorem.

1. Find the multiplicative factors of the leading coefficient and the constant term.
2. Find all the values of frac(p,q), where p is a factor of the constant term and q is a factor of the leading coefficient. Consider both positive and negative options.
3. Find out which of the possible zeros in $fleft(fracpqright$ are true zeros.

For the function f(x), write down all the possible rational zeros: $f(x)=2x4-5x3 x2-4$.

To determine the rational zeros of the function f(x), where x is a real number, use the Rational Zero Theorem.

Find the rational zeros of the function f(x)=x3-3x2-6x8 using the Rational Zero Theorem.

Another useful theorem for analyzing polynomial equations is the Factor Theorem. It reveals the relationship between the factors and the zeros of a polynomial. You may remember that $fleft(xright)=left(x-kright)qleft(xright) r$ from the Division Algorithm.

The remainder r is $fleft(kright)=0$ if and only if k is zero, and $fleft(xright)=left(x-kright)qleft(xright) 0$ or $fleft(xright)=left(x-kright)qleft(xright)$ if k is nonzer

Note that x - k is a factor of $fleft(xright] in this notation. Since [latex]x-k$ is a factor of $fleft(xright]$, we know that if k is a zero of $fleft(xright]$, then x-k is a factor of $fleft(xright]$.

The remainder of the division algorithm $fleft(xright)=left(x-kright)qleft(xright) r$ is also zero if $x-k$ is a factor of $fleft(xright)$. This demonstrates that k is a digit zero.

The Factor Theorem is a set of these two conclusions. In the complex number system, a polynomial of degree n will have exactly n zeros, as we shall see. By applying the Factor Theorem, we can break down any polynomial into its product of n factors. Once the polynomial has been fully factored, finding its zeros is a breeze.

The Factor Theorem states that an integer k is a zero of the function f(x) if and only if the function is a factor of itself.

### How to: Apply the Factor Theorem to factor a polynomial of degree 3 when given a factor and the degree of the polynomial.

1. Polynomial division by $left(x-k)right$ is an example of synthetic division.
2. Verify that there is no decimal remainder.
3. Multiply $left(x-k)right$ by the quadratic quotient to get the polynomial.
4. Factor the quadratic if you can.
5. Multiply the factors to form the polynomial.

To prove this, we need to establish that $left(x 2right)$ is a factor of $x3-6x2-x 30$. Figure out the missing variables. The zeros of the polynomial can be calculated using the factors.

Given that $left(x - 2right)$ is a factor of the polynomial, you can use the Factor Theorem to determine its zeros.

By using the Rational Zero Theorem, we can reduce the number of rational zeros for a polynomial function to a manageable number. Once this is accomplished, we can use synthetic division indefinitely to find all of a polynomial function's zeros.

### Synthetic division can be used to locate the zeros of a polynomial function, such as $f$.

1. You can find all of the rational zeros of the function by applying the Rational Zero Theorem.
2. You can judge the quality of a candidate for the zero by synthetic division into the polynomial. The candidate is a zero if the remainder is 0. Discard the candidate if the difference is greater than zero.
3. The result of synthetic division can be used to redo step 2. If you can, keep going until you get a quadratic quotient.
4. Determine where the zeros of the quadratic are. Factoring and the quadratic formula are two approaches to solving quadratics.

Determine the values of x such that f(x)=4x3 - 3x - 1 has no real roots.

We will now examine a theorem that addresses the question of how many complex zeros a polynomial function has now that we know how to find its rational zeros. Every polynomial function, according to the Fundamental Theorem of Algebra, has at least one complex zero. The solution of polynomial equations relies on this theorem.

Consider the polynomial function f of degree 4, and let fleft(xright)=0. According to the Fundamental Theorem of Algebra, there exists at least one complex solution, which we will denote by $c_1$. The Factor Theorem states that the product $fleft(xright)$ of $x-c_text1$ and a polynomial quotient is equal to $fleft(xright)$. Since $x-c_text1$, the polynomial quotient will be of degree three since it is linear. Now we divide a polynomial by a third-degree polynomial using the Fundamental Theorem of Algebra. At the very least, it will have a complex zero, so we'll refer to it as $c__text2$. A new polynomial quotient of degree two can be written as the product of the polynomial quotient and the expression $x-c_text2$. In order to find all of the zeros, the Fundamental Theorem of Algebra must be repeatedly applied. There will be a total of four, and the product of each will be $fleft(xright] If [latex]f(x)$ is a polynomial of degree $n>0$, then $f(x)$ has at least one complex zero.

If f(x) is a polynomial of degree $n>0$, and a is a non-zero real number, then we can use this theorem to prove that f(x) is exactly divisible by n linear factors.

We can write the polynomial as

$f\left(x\right)=a\left(x-{c}_\right)\left(x-{c}_\right)…\left(x-{c}_{n}\right)$

complex numbers $c__1, c__2,..., c__n$ So, if we permit multiplicities, we find that $fleft(xright] has n roots. Does every polynomial have at least one imaginary zero No It is not always the case that an imaginary number is complex. Complex numbers include all real numbers. Determine the values of x such that [latex]f(x)=3x3 9x2 x 3$ have no integer solutions.

Determine the values of x for which the function f(x)=2x3 5x2 -11x4 has no solutions.

## Factoring by linear combinations and the Descartes' rule of signs

Allowing for multiplicities in the set of complex numbers leads to a crucial implication of the Fundamental Theorem of Algebra: a polynomial function of degree n will have n zeros. That's to say, we can break down the polynomial into n smaller functions. According to the Linear Factorization Theorem, the number of factors in a polynomial function is proportional to the degree of the function, and each factor is of the form (x - c), where c is a complex number.

Assume that a bitext, bne 0, is a zero of the polynomial function f(x), where the coefficients are real numbers. Then, using the Factor Theorem, we know that $x-left(a biright)$ is a factor of $fleft(xright). The coefficients of f must be real numbers if and only if [latex]x-left(a-bi)right$. This is the case because multiplying any other factor by $x-left(a-bi)$ will result in a product with imaginary parts. The only way to get rid of the imaginary parts and get real coefficients is to multiply by conjugate pairs. That is, if a polynomial function f with real coefficients has a complex zero $a bi$, then $a-bi$ must also be a zero of $fleft(xright]$. The term for this is the Theorem about Complex Conjugates.

Each factor of a polynomial function will take the form $left(x-c)right$ where c is a complex number, as stated by the Linear Factorization Theorem.

The complex conjugate of a complex zero of the form $a bi$ is also a zero if and only if the polynomial function f has real coefficients.

### How To: Apply the Linear Factorization Theorem to determine the polynomial function f given its zeros and a point on its graph represented by $left(ctext, f(c)right] 1. Construct the linear factors of the polynomial using the zeros. 2. For a larger polynomial, multiply the linear factors by themselves. 3. To find the leading coefficient, plug in [latex]left(c,fleft(c)right]$.
4. Simplify

Find a fourth degree polynomial with real coefficients that has zeros of –3, 2, i, such that $f\left(-2\right)=100$

Does 2 - 3i also need to be a zero of the polynomial if 2 3i is given as a zero? (The polynomial has real coefficients.)

Yes In the case where an imaginary zero is given for a polynomial with real coefficients, the conjugate must also be a real zero.

Determine a real-coefficient polynomial of degree 3 with the zeros 5 and -2i, satisfying the equation $fleft(1right)=10$.

For any given polynomial function, it is easy to count the number of positive and negative real zeros. Descartes' Rule of Signs states that there is a correlation between the number of sign changes in $fleft(xright]$ and the number of positive real zeros if the polynomial is written in descending order.

In the same way that the number of negative real zeros is proportional to the number of sign changes in $fleft(-xright$, so is the number of sign changes in $fleft(-x Descartes's Rule of Signs states that for any polynomial function with real coefficients, such as [latex]fleft(xright)=a_nxn a_n - 1xn - 1... a_1x a_0$, then:

• Either the number of sign changes in $fleft(xright]$ is an even number, or the number of positive real zeros is an even number less than the number of sign changes.
• Either the number of sign changes in $fleft(-xright]$ is an even number, or the number of negative real zeros is an even number.

Apply Descartes' Rule of Signs to $fleft(xright]=-x4-3x3 6x2-4x12$ to find the number of positive and negative real zeros.

Determine the maximum number of positive and negative real zeros for f(x)=2x4-10x3 11x2-15x12 using Descartes' Rule of Signs. Check the function's real positive and real negative zeroes using a graph.

### Polynomial Equations and Their Practical Applications

At this point, we have covered a wide range of methods for addressing polynomial equations. Let's apply these methods to the bakery conundrum introduced at the beginning of the article.

A new bakery has plain and themed sheet cakes for all your celebration needs. The cake shop requests that a miniature cake has a volume of 351 cubic inches. The cake is a regular rectangular shape. The cake's height should be one-third of its width, and the length should be four inches longer than the width. What should the cake pan's measurements be?

The minimum volume for a rectangular, solid shipping container is 84 cubic meters. The client specifies to the manufacturer that the container's length must be one meter longer than the width and the height must be one meter greater than twice the width in order to accommodate the contents. How big should the storage unit be?

## Important Ideas

• Determine the remainder of the polynomial $fleft(xright] when it is divided by [latex]x-k$ to get $fleft(kright]. • If [latex]left(x-k)right$ is a factor of $fleft(x-right]$, then $kright$ is a zero of $fleft(x-right]. • A polynomial function with integer coefficients will have rational zeros that are proportional to the product of a factor of the constant term and a factor of the leading coefficient. • The factors of the constant term are the rational zeros when the leading coefficient is 1. • Zeros of a polynomial function can be located using synthetic division. • Every polynomial function has at least one complex zero, as proven by the Fundamental Theorem of Algebra. • At least one complex zero exists for every non-zero degree polynomial function. • In the presence of multiplicities, the number of factors in a polynomial function equals the degree of the function. To express this, we will write each factor as [latex]left(x-c)right$, where c is a complex number.
• A polynomial function has the same number of positive real zeros as the number of times it changes signs, or an even smaller number.
• The number of negative real zeros of a polynomial function is exactly the same as, or exactly one less than, the number of sign changes of $fleft(-xright]$.
• Various situations in the real world can be modeled by polynomial equations. The use of synthetic division to solve the equations is recommended.

## Glossary

Descartes's "Sign Theory" a formula that, given the number of sign changes of $fleft(xright)$ and $fleft(-xright)$, yields the maximum number of positive and negative real zeros, respectively. Use of the Factor Theorem If and only if $left(x-k)right$ is a factor of $fleft(x)right$, then k is a zero of the polynomial function $fleft(x)right$. Algebra's Central Theorem Every non-zero degree polynomial function has at least one complex zero. Factorization Theorem, Linear Taking into account the multiplicities, the number of factors of a polynomial function is equal to its degree, and each factor is of the form $left(x-cright]$, where c is a complex number. Theorem about rational zeros If p is a factor of the constant term and q is a factor of the leading coefficient, then the rational zeros of the polynomial function have the form $fracpq$. Theorem for a Remainder the remainder obtained by dividing a polynomial $fleft(xright] by [latex]x-k$ is equal to [latex]fleft(kright].
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